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How does multiplying 2 segments give a surface?

  1. Oct 30, 2003 #1
    how did they ever think of that?
    what's the reasoning behind?

    sorry if it's stupid

    bonus question: in a formula such as E=mc2, what is the significance of the square? what exactly does a multiplication such as these, or a fraction mean in physics? doest the square hold a special meaning, compared to the "common" multiplication?

    thanks
     
  2. jcsd
  3. Oct 30, 2003 #2

    HallsofIvy

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    Perhaps you could tell us what YOU mean by "multiplying two segments".

    No, there are no "special meanings" in mathematics. What you see is what you get.
     
  4. Oct 30, 2003 #3

    Hurkyl

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    Are you talking about something like

    [0, 1]x[3, 4]

    ?

    (this is called a Cartesian product)
     
  5. Oct 30, 2003 #4

    Integral

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    Physically consider the units of various quanities.

    For example area has units of m2 (this is metric of course) that is simply m X m or as you think of it the product of 2 intervals.

    Now what is the siginifance of c2 in E = mc2

    E is of course Energy, energy has units of kg m2/s2


    Which if you look at is mass times velocity squared. The c has to be squared for the equation to make any sense.
     
  6. Oct 31, 2003 #5
    thanks, but ...

    i'll understand if neither of you wants to waste any more time with me
    still...
    in the first case (geometry):
    the "interval" you mention is unidimensional; its "width" is zero (that's what I meant by multiplying two "segments"); how do you reach the conclusion that you have to multiply two lengths to get an area, since we are not talking of a band gliding sidewards on another band (sweeping), but of two unidimensional segments? units have nothing to do with this, since they are a result of the reasoning I have been inquiring about.

    in the next case (physics):
    about square meters we discussed above...
    and it's the only "graspable" notion
    my question was more in line with what does a square second mean

    thanks again
     
  7. Oct 31, 2003 #6

    HallsofIvy

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    Oh, what the heck! I wouldn't be here if I didn't want to waste time! (Hey, it better than cleaning my room!)

    Multiplying two lengths is NOT the same as "multiplying two segments". Segments are geometric objects- lengths are numbers.
    The answer is that we DEFINE our unit area in terms of a square one unit on a side. The question is then how many of those we can fit into the figure (a rectangle, say). You learned way back in elementary school that if you can fit "l" things onto each row and "w" rows into a box, then there are l*w things altogether. THAT'S the idea of "length times length".

    On the other hand, when you are talking about "E= mc2" where c is in, say, m/sec., so that c2 is in m2/sec2 or "F= ma" where a (acceleration) is measured in m/sec2, you are NOT talking about "seconds squared"- the sec2 is in the denominator so it is really "per second per second".
     
  8. Nov 3, 2003 #7
    Bear with me a little longer. :)
    I understand the situation where time is the denominator and the significance of time squared is "per second, per second". Therefore I understand the expression of acceleration m/sec2; what I still don't get is the significance of m2/sec2; where does the surface fit in? I mean how does the square of the speed, a vector, therefore unidimesnional, becomes a surface? what is the significance of "area per second per second" and how does it relate to speed?

    Silly of me, but you know how it is: suddenly you realize you have taken some things for granted and it's like you've seen them for the first time, and you need to check with someone.

    Now, the geometry part...My question was triggered by the reading of
    Newton's Principa...
    (SECTION I.
    Of the method of first and last ratios of quantities, by the help whereof we demonstrate the propositions that follow.
    ...
    It may also be objected, that if the ultimate ratios of evanescent quantities are given, their ultimate magnitudes will be also given: and so all quantities will consist of indivisibles, which is contrary to what Euclid has demonstrated concerning incommensurables, in the 10th Book of his Elements. But this objection is founded on a false supposition. For those ultimate ratios with which quantities vanish are not truly the ratios of ultimate quantities, but limits towards which the ratios of quantities decreasing without limit do always converge; and to which they approach nearer than by any given difference, but never go beyond, nor in effect attain to, till the quantities are diminished in infinitum.)

    Q: Does your explanation mean that if we choose to define our area unit as a square, there will always be, to the limit, a smaller square, so that the empiric geometric construction also “holds” in an abstract Euclidean (continuous) space (and vice versa)?
     
  9. Nov 3, 2003 #8

    joc

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    gabriel - i think that the most satisfactory answer to your question would lie in the einstein's actual derivation of the equation that gives E = mc^2. btw c isn't a vector here; it's a scalar constant that is found to be the factor for the conversion of a numerical mass into the equivalent quantity of energy.
     
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